Question

Drive the prism formula, $n_{12}=\frac{Sin\frac{(A+\delta_{m})}{2}}{Sin\frac{A}{2}}$
Draw the graph showing the variation of the angle of deviation with angle of incidence, through a prism.

Answer 1

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In the figure attached,

 δ = i1–r1+ i2−r2δ

= i1–r1+i2−r2  ….... (1)

δ = i1+i2–(r1+r2)δ

=i1+i2–(r1+r2)

Again, in quadrilateral ALOM,

∠ALO + ∠AMO = 2rt∠s

[Since, ∠ALO = ∠AMO = 90º]

So,

∠LAM +∠LOM = 2rt∠s

[Since, Sum of four ∠s of a quadrilateral = 4 rt∠s] ….... (2)

Also in ∠LOM,

∠r1+∠r2+∠LOM

= 2rt∠s∠r1+∠r2+∠LOM=2rt∠s  …... (3)

Comparing (2) and (3),

we get,

∠LAM =∠r1+∠r2∠LAM=∠r1+∠r2

A=∠r1+∠r2A=∠r1+∠r2

Using this value of ∠A,

equation (1) becomes,

δ=i1+i2−Aδ=i1+i2−A

or

 i1+i2=A+δi1+i2=A+δ  …... (4)

Now,

In the minimum deviation position,

 ∠i1=∠i2∠i1=∠i2

and so

 ∠r1=∠r2=∠r∠r1=∠r2=∠r (say)

Obviously,

∠ALM = ∠LMA = 90º – ∠r

Thus,

AL = LM

and so

LM l l BC

Hence,

the ray which suffers minimum deviation possess symmetrically through the prism and is parallel to the base BC.

Since for a prism,

∠A=∠r1+∠r2∠A=∠r1+∠r2

So,

A = 2r

(Since, for the prism in minimum deviation position, ∠r1=∠r2=∠r∠r1=∠r2=∠r)

or

r = A/2 …...(5)

Again, 

i1+i2=A+δi1+i2=A+δ

or 

i1+i1=A+δmi1+i1=A+δm 

(Since, for the prism in minimum deviation position, i1=i2andδ=δmi1=i2andδ=δm)

2i =A+δm2i1=A+δm

or 

i1=(A+δm)/2i1  …... (6)

Now 

µ=sini1/ sinr1

=sini1/ sinr

HENCE,

$n_{12}=\frac{Sin\frac{(A+\delta_{m})}{2}}{Sin\frac{A}{2}}$

This is the Relation Between Refractive Index and Angle of Minimum Deviation

Answer 2

In the figure attached,

 δ = i1–r1+ i2−r2δ

= i1–r1+i2−r2  ….... (1)

δ = i1+i2–(r1+r2)δ

=i1+i2–(r1+r2)

Again, in quadrilateral ALOM,

∠ALO + ∠AMO = 2rt∠s

[Since, ∠ALO = ∠AMO = 90º]

So,

∠LAM +∠LOM = 2rt∠s

[Since, Sum of four ∠s of a quadrilateral = 4 rt∠s] ….... (2)

Also in ∠LOM,

∠r1+∠r2+∠LOM

= 2rt∠s∠r1+∠r2+∠LOM=2rt∠s  …... (3)

Comparing (2) and (3),

we get,

∠LAM =∠r1+∠r2∠LAM=∠r1+∠r2

A=∠r1+∠r2A=∠r1+∠r2

Using this value of ∠A,

equation (1) becomes,

δ=i1+i2−Aδ=i1+i2−A

or

 i1+i2=A+δi1+i2=A+δ  …... (4)

Now,

In the minimum deviation position,

 ∠i1=∠i2∠i1=∠i2

and so

 ∠r1=∠r2=∠r∠r1=∠r2=∠r (say)

Obviously,

∠ALM = ∠LMA = 90º – ∠r

Thus,

AL = LM

and so

LM l l BC

Hence,

the ray which suffers minimum deviation possess symmetrically through the prism and is parallel to the base BC.

Since for a prism,

∠A=∠r1+∠r2∠A=∠r1+∠r2

So,

A = 2r

(Since, for the prism in minimum deviation position, ∠r1=∠r2=∠r∠r1=∠r2=∠r)

or

r = A/2 …...(5)

Again, 

i1+i2=A+δi1+i2=A+δ

or 

i1+i1=A+δmi1+i1=A+δm 

(Since, for the prism in minimum deviation position, i1=i2andδ=δmi1=i2andδ=δm)

2i =A+δm2i1=A+δm

or 

i1=(A+δm)/2i1  …... (6)

Now 

µ=sini1/ sinr1

=sini1/ sinr

HENCE,

n_{12}=\frac{Sin\frac{(A+\delta_{m})}{2}}{Sin\frac{A}{2}}n12=Sin2ASin2(A+δm)

This is the Relation Between Refractive Index and Angle of Minimum Deviation